Does Tarzan's Vine Hold During His Daring Swing?

[matt57017]

Homework Statement

Tarzan, who weighs 678 N, swings from a cliff at the end of a convenient vine that is 16.0 m long (see the figure). From the top of the cliff to the bottom of the swing, he descends by 3.2 m. A) If the vine doesn't break, what is the maximum of the tension in the vine? B) The vine will break if the force on it exceeds 928.6 N. Does the vine break? If yes, at what angle does it break

Homework Equations

F=ma, Emec=Δk+ΔU, F=mv²/r

The Attempt at a Solution

A) here is my thought process to this problem, if i am wrong in my approach please correct me...tarzan starts at let's say point 1 with no potential energy, but has kinetic energy as he begins his swing. At the bottom of his swing (lowest point) will be where the max tension will occur. So, F=ma becomes F=mv²/r and you also have mgh=1/2mv² because he drops 3.2m so by figuring out the GPE and equating that to the final KE (bottoms of swing) you will derive the velocity at the bottom of the swing. i end up with F=mg(1+2h/r) and get 949.2 N which is correct.
B) i have no idea how to find the angle where it breaks at or even how to derive the formula...this is where i am hoping someone can guide me in the right direction (how to analyze the problem).

Thank you for your time!

 

[Yosty22]

tarzan starts at let's say point 1 with no potential energy

I believe that is an incorrect statement. I don't have the picture to go off of, but in the problem, it says he descends 3.2m, which means he would have had to have started at least 3.2m off the ground, if not more. In this case, he would have potential energy.

 

[matt57017]

Yosty,

that is true my fault. i have attached a picture for further clarification.

 

[Fluxthroughme]

I believe that is an incorrect statement. I don't have the picture to go off of, but in the problem, it says he descends 3.2m, which means he would have had to have started at least 3.2m off the ground, if not more. In this case, he would have potential energy.

You can define potential energy to be zero wherever you'd like, provided you're alright with using negatives.